Maths Olympiad Prep

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Combinatorics Difficulty 5.3 AIME, harder Find the answer

3. Five friends make the following statements, respectively.

"No matter which one of us is chosen, the other 4 lie".

"No matter which one of us is chosen, the other 4 tell the truth".

"No matter which one of us is chosen, there is another one who tells the truth".

"There is one of us such that every other tells the truth".

"There is one of us such that every other lies".

Which of the following statements can be deduced from the above?

(A) Exactly 1 tells the truth.

(B) Exactly 2 tell the truth.

(C) Exactly 3 tell the truth.

(D) Exactly 4 tell the truth.

(E) It is not possible to determine the number of those who tell the truth.

Solution

3. Evaluate with intermediate scores partial or only partially correct solutions.

For example, in the case of exercise 16, assign (the suggested scores are not cumulative):

- 1 point for correctly drawing the figure
- 4 points for proving that triangle ABDA B D is isosceles
- 5 points for also observing the equality of angles DA^H,BA^HD \widehat{A} H, B \widehat{A} H
- 7 points for proving that E,HE, H belong to the circle with diameter ACA C
- 9 points for recognizing BA^HB \widehat{A} H as an angle at the circumference that subtends arc AHA H
- 12 points obviously, for concluding the proof.

For example, in the case of exercise 17, assign (the suggested scores are not cumulative):

- 1 point for noting the solutions with n=1n=1 and n=2n=2
- 2 points for noting that it is possible to simplify a factor of 3
- 3 points for making both of the above observations
- 4 points for finding all solutions (even without simplifying the factor of 3)
- 5 points for simplifying the factor of 3, finding all solutions, and conjecturing (without proof) that there are no others
- 5 points for simplifying the factor of 3 and attempting to use notable products to factorize the numerator (at least in one case)
- 8 points for correctly solving one of the two cases (mm even or mm odd)
- 9 points for correctly solving one of the two cases and providing the correct conjecture for the other case
- 12 points obviously, for concluding the proof.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.